minimum of y

The solution is only true for $x > 0$ .

$\begin{aligned} y &= x^2 +8x +\frac {64}{x^3} \\ \frac {dy}{dx}&= 2x+8-\frac {192}{x^4} & \small \color{#3D99F6} \text{Putting }\frac {dy}{dx}=0 \end{aligned}$

$\begin{aligned} 2x+8 -\frac {192}{x^4}&=0 \\ \implies x &= 2 \\ \frac {d^2y}{dx^2}&= 2+\frac {768}{x^5} \\ \frac {d^2y}{dx^2}\bigg|_{x=2} &= 2+\frac {768}{32} > 0 \\ \implies \min(y) &=y(2) \\ & = 2^2+8(2)+\frac {64}{2^3} \\ &=\boxed{28} \end{aligned}$